Let $f(x)=\dfrac{1}{2}x^4+x^3-6x^2$. For what values of $x$ does the graph of $f$ have a point of inflection? Choose all answers that apply: Choose all answers that apply: (Choice A) A $x=-2$ (Choice B) B $x=0$ (Choice C) C $x=1$ (Choice D) D $f$ has no points of inflection.
Answer: We can find the inflection points of the graph of $f$ by looking for the intervals where its second derivative $f''$ is positive/negative. This analysis is very similar to finding minimum/maximum points, only instead of analyzing $f'$, we are analyzing $f''$. The second derivative of $f$ is $f''(x)=6(x-1)(x+2)$. $f''(x)=0$ for $x=-2,1$. Since $f''$ is a polynomial, it's defined for all real numbers. Therefore, our possible inflection points are $x=-2$ and $x=1$. Our possible inflection points divide the number line into three intervals: $\llap{-}6$ $\llap{-}5$ $\llap{-}4$ $\llap{-}3$ $\llap{-}2$ $\llap{-}1$ $0$ $1$ $2$ $3$ $4$ $5$ $x< \llap{-}2$ $\llap{-}2<x<1$ $x>1$ Let's evaluate $f''$ at each interval to see if it's positive or negative on that interval. Interval $x$ -value $f''(x)$ Verdict $x<-2$ $x=-3$ $f''(-3)=24>0$ $f$ is concave up $\cup$ $-2<x<1$ $x=0$ $f''(0)=-12<0$ $f$ is concave down $\cap$ $x>1$ $x=2$ $f''(2)=24>0$ $f$ is concave up $\cup$ We can see that the graph of $f$ changes concavity at both $x=-2$ and $x=1$. In conclusion, these are the values of $x$ where the graph of $f$ has a point of inflection: $x=-2$ $x=1$